Approximative Compactness and Radon-Nikodym Property in Nearly Dentable Banach Spaces and Applications

Abstract

Authors define nearly dentable Banach space. Authors study Radon-Nikodym property, approximative compactness and continuity metric projector operator in nearly dentable space. Moreover, authors obtain some examples of nearly dentable space in Orlicz function spaces. Finally, by the method of geometry of Banach spaces, authors give important applications of nearly dentability in generalized inverse theory of Banach space.

1. Introduction

Let be a real Banach space. and denote the unit sphere and the unit ball, respectively. By we denote the dual space of . Let , and denote the set of natural numbers, reals, and nonnegative reals, respectively. and . By   denote that is weakly convergent to . denote that is convergent to . The weak topology of is denoted by or . The topology of is denoted by or . denote closed hull of (weak closed hull, closed hull closed hull). dist denote distance of and .

The study of geometric and topological properties of unit balls of Banach spaces plays a central rule in the geometry of Banach spaces. Almost all properties of Banach spaces, such as convexity, smoothness, nonsquareness, and the Radon-Nikodym property, can be viewed as properties of the unit ball. We should mention here that there are many topics studying behavior of ball collections. For example, the Mazur intersection property, the packing sphere problem of unit balls, the measure of noncompactness with respect to topological degree, and the ball topology have also brought great attention of many mathematicians. In 2006, Cheng defined ball-covering property of Banach space (see [1]). In [2], Chen and Cheng proved analytical characterizations of Mazur intersection property.

Starting with a different viewpoint, a notion of nearly dentable space is introduced by Shang et al. in [3].

Definition 1. Banach space is called nearly dentable if, for any and any open set , we have and , where .

By James theorem, it is easy to see that nearly dentable space is reflexive.

Definition 2. A nonempty subset of is said to be approximatively compact if for any and any satisfying , then has a subsequence converging to an element in . is called approximatively compact if every nonempty closed convex subset of is approximatively compact.

Let us present the history of the approximative compactness and related notions. This notion has been introduced by Jefimow and Stechkin in [4] as a property of Banach spaces, which guarantees the existence of the best approximation element in a nonempty closed convex set for any . In [3] it was established (without proof) that the approximative compactness of in a rotund Banach space gives also the continuity of the projector operator . Fully rotund Banach spaces were defined by Fan and Glicksberg in [5]. In [6] it has been proved that fully rotund Banach spaces are approximately compact. In [7] it has been proved that approximative compactness of a Banach space means that is reflexive and it has the Kadec-Klee property (called also the property ). In 2007, Chen et al. (see [8]) proved that a nonempty closed convex of a midpoint locally uniformly rotund space is approximately compact if and only if is a Chebyshev set and the metric projection operator is continuous. In 1972, Oshman (see [9]) proved that the metric projection operator is upper semicontinuous if is approximatively compact subset of a Banach space .

Let be a nonempty open convex subset of and a real-valued continuous convex function on . is said to be Gateaux differentiable at the point in if the limit exists for all . When this is the case, the limit is a continuous linear function of , denoted by . If the difference quotient in converges to uniformly for in the unit ball, then is said to be Fréchet differentiable at . is called a weak Asplund space (Asplund space) or said to have the weak Asplund property if, for every and as above, is “generically” Gateaux (Frechet) differentiable; that is, there exists a dense subset of such that is Gateaux (Frechet) differentiable at each point of .

The paper is organized as follows. In Section 1, authors define nearly dentable Banach space and some necessary definitions and notations are collected. In Section 2, authors prove that if is a nearly dentable space, then is an Asplund space if and only if is a separable subset of , where is a separable closed subspace of . In Section 3, authors prove that the following statements are equivalent: every hyperplane of is approximatively compact; if is a open convex set, then is approximatively compact;    is nearly dentable space and is compact for any . Moreover, authors also prove that if is nearly dentable space, then for any open convex set metric projector operator is upper semicontinuous; For any hyperplane metric projector operator is upper semicontinuous. In Section 4, authors prove that there exists a Banach space such that is not nearly dentable Banach space. Moreover, authors also prove that there exists a Banach space such that is nearly dentable Banach space and not nearly dentable Banach space. The main results in this paper have important applications to the ill-posed equation theory and the generalized inverse theory. In Section 5, authors give an important application of nearly dentability in generalized inverse theory of Banach space.

Definition 3. is called nearly dentable space, if, for any and open set , we have .

Let us recall some definitions and some lemmas which will be used in the further part of the paper.

Definition 4 (see [8]). A subset is said to be proximinal if for all . is said to be semi-Chebyshev if is at most a singleton for all . is said to be Chebyshev if it is proximinal and semi-Chebyshev.

Definition 5. A Banach space is said to have the Radon-Nikodym property whenever if is a nonatomic measure space and is a vector measure on with values in which is absolutely continuous with respect to and has bounded variation; then there exists such that, for any , It is well known that is reflexive if and only if each closed convex subset of is proximinal. It is well known that is Asplund space if and only if has the Radon-Nikodym property.

Definition 6 (see [10]). Set-valued mapping is called upper semicontinuous at , if, for each norm open set with , there exists a norm neighborhood of such that for all in . is called lower continuous at , if, for any and any in with , there exists such that as . is called continuous at , if is upper semicontinuous and is lower continuous at .

Definition 7. is called hyperplane, if there exist and such that .

Lemma 8 (see [11]). Suppose that is a Banach space. Then the following statements are equivalent:(1)is Asplund space.(2)If is a separable closed subspace of , then is a separable space.(3) has the Radon-Nikodym property.

Lemma 9 (see [12]). Let and be not an emptyset for all . Assume that for some countable set and some . Then is the closed linear span of .

2. Nearly Dentability and Radon-Nikodym Property in Banach Space

Theorem 10. Suppose that is a nearly dentable space. Then the following statements are equivalent:(1) is a Asplund space.(2) has the Radon-Nikodym property.(3)If is a separable closed subspace of , then is a separable space.(4)If is a separable closed subspace of , then is a separable subset of .

Proof. Let be a separable closed subspace of and be a dense subset of . Then, for any , we have In fact, suppose that . Then, for any , there exist and such that , where . This implies that Since is arbitrary, we obtain that This implies that there exists such that . Hence we define an open set Since is a nearly dentable space, we have . Since is a compact set, by the separation theorem of locally convex space, there exists such that Since the sequence is a dense subset of , we may assume without loss of generality that as . Pick . Then Moreover, we may assume without loss of generality that if , then . Since is compact, there exists such that is a accumulation point of . This implies that . Then . Therefore, by formula (6), we obtain that Since and , by formula (6), we have , which contradicts . Hence formula (2) is true. Since is a separable subset of , we obtain that there exists a countable dense subset of for each . Then is a countable set. Pick . Therefore, by formula (2), we obtain that there exist and such that for any . Hence, for every , we obtain that for any . Since and is a countable dense subset of , we obtain that for any . This implies that for any . Noticing that , we have By Lemma 9, we obtain that , and so is separable.
This means that if is a separable closed subspace of , then is a separable space. By Lemma 15, we obtain that is Asplund space and has the Radon-Nikodym property.
. Since has the Radon-Nikodym property, we obtain that if is a separable closed subspace of , then is a separable space. Hence is a separable subset of .
Therefore, by Lemma 8, we obtain that Theorem 10 is true, which completes the proof.

3. Nearly Dentability and Approximative Compactness and Continuity of Metric Projector Operator in Banach Spaces

Theorem 11. Suppose that is a Banach space. Then the following statements are equivalent:(1)Every hyperplane of is approximatively compact.(2)If is a open convex set, then is approximatively compact.(3) is nearly dentable space and is compact for any .

Proof. . (a) Let as , where and . Since is compact, there exists such that is a accumulation point of . Hence there exists a net such that and . We next will prove that the sequence is relatively compact. The proof requires considering few cases separately.
Case I . Suppose that is not relatively compact. Then there exists a subsequence denoted again by such that does not contain convergent subsequence. This implies that every point of is not accumulation point of . It is easy to see that for any , there exists such that . Put Then, it is easy to see that . Moreover, since is a compact convex set, we obtain that is a compact convex set. Since is a nearly dentable space, we obtain that . By separation theorem, there exist and such that From the previous proof, there exists such that . By , we have Therefore, by formulas (14) and (15), we obtain that . This implies that is impossible, a contradiction. Hence is relatively compact.
Case II . From the previous proof, is relatively compact. Since is compact, we obtain that is relatively compact. Noticing that , we obtain that is relatively compact.
Let be a open convex set and . Then, for any . we may assume without loss of generality that . We next will prove that if as , then the sequence has a subsequence converging to an element in , where . Since is a closed set, by proof of theorem 2.1 of [13], we obtain that is a proximinal set and . Pick . Then . In fact, suppose that . Then there exist and such that and . Since we obtain that a contradiction. Hence Noticing that and separation theorem of locally convex space, there exists such that This implies that Since , we obtain that This means that the inequality holds. Therefore Hence and as . Furthermore, we have which shows that . From the previous proof, we obtain that the sequence is relatively compact. Noticing that as , we obtain that the sequence is relatively compact. This implies that the sequence has a subsequence converging to an element in . Hence we obtain that is approximatively compact.
. Let be a hyperplane. Then it is easy to see that and are closed convex sets. Moreover, it is easy to see that if , then and if , then . Since and are approximatively compact, we obtain that is approximatively compact.
. For any , we define a hyperplane . Let . Then . Hence has a subsequence converging to . Moreover, it is easy to see that is a closed set. Hence is compact. We claim that for any open set . In fact, suppose that . Then there exists such that dist as . Since is compact, there exists such that . Since is compact, without loss of generality, we may assume that is a Cauchy sequence. Let as . Then . Hence there exists such that . Moreover, we have Therefore, , when is large enough, which contradicts . Hence, for any open set , we have . We next will prove that there exists such that whenever . Otherwise, there exists such that . It is easy to see that . Let . Then it is easy to see that . Hence we define a hyperplane . Then it is easy to see that dist. Hence, for any , there exists a net such that . Moreover, it is easy to see that . This implies that . Hence we obtain that is a closed set. By the proof of Theorem 2.1 of [13], we know that if is a closed set, then is a proximinal set. Hence there exists such that . Pick . Then . Since we obtain that Therefore, by and formula (27), we obtain that Since every hyperplane of is approximatively compact, we obtain that has a subsequence converging to an element in . By formula (26), we obtain that has a subsequence converging to a element in . Hence we may assume without loss of generality that . Then . By and , we have . This implies that . Then dist as . By and , we obtain that dist as . Moreover, we have proved that for any open set , a contradiction. This implies that there exists such that whenever . Hence, for any , we have This implies that . Therefore, by the arbitrariness of , we have . Hence is nearly dentable space. This completes the proof.

Theorem 12. Suppose that is nearly dentable space. Then, (1)for any open convex set , metric projector operator is upper semicontinuous;(2)for any hyperplane , metric projector operator is upper semicontinuous.

Proof. Let be a open convex set and . Suppose that metric projector operator is not upper semicontinuous. Then we without loss of generality may assume that metric projector operator is not upper semicontinuous in orgin. Hence there exist a sequence and an open set such that , and as . This implies that there exists such that . We will derive a contradiction for each of the following two cases.
Case I. Let . Then, by , we have dist as . Moreover, by , we have as . Therefore, This implies that as . Hence there exists such that , a contradiction.
Case II. Let . Then , , and . Therefore, by and the separation theorem of locally convex space, there exists such that . This implies that Therefore, by , we obtain that for any . Since , we obtain that . Since the distance function is continuous, we obtain that Moreover, we have Therefore, by formulas (32) and (33), we have . By , we get that every subsequence does not converge to point of . Moreover, by , we have . Hence, for any , there exists such that . Since is closed convex set and , we obtain that for any , there exists such that dist. This implies that . Noticing that we obtain that, for any , there exists such that Therefore, by and , we have Let us define sequence , where Then Noticing that , we obtain that . This implies that . Since , we have Hence there exists such that whenever . Therefore, by formula (35), we have . Hence if , then Since and , we have . Put Then, by formulas (40) and (41), we have . Moreover, by the arbitrariness of , we have Put Then, by formulas (42) and (43), we have .
Since the distance function is continuous, we have as . Moreover, we have as . This implies that as . Therefore, by as , we obtain that Since , we obtain that as . By we have as . Since is compact, there exists such that is a accumulation point of . Hence there exists a net such that and . Since , we obtain that Let . Then, by and , we have . This implies that and . Therefore, This implies that . Since is a nearly dentable space, we obtain that . Since is compact, it is easy to see that is compact. Moreover, since is closed convex set, by the separation theorem of locally convex space, there exist and such that Therefore, by , we obtain that . Noticing that , we obtain that Therefore, by formulas (48)-(49) and , we have , which contradicts . Hence the metric projector operator is upper semicontinuous.
By the proof of Theorem 11, it is easy to see that is true. This completes the proof.

4. Some Examples of Nearly Dentable Space in Orlicz Function Space

In the section, authors prove that there exists a Banach space such that is not a nearly dentable Banach space. Moreover, authors also prove that there exists a Banach space such that is a nearly dentable Banach space and not a nearly dentable Banach space. First, we give the definition of Orlicz function space.

Definition 13. is called an function if it has the following properties: (1) is even, convex, and ;(2) for all ;(3) and .

Let denote the right derivative of at and let be the generalized inverse function of defined on by . Then we call the complementary function of . It is well known that there holds the Young inequality and or . Moreover, it is well known that and are complementary to each other. A continuous function is called strictly convex if for all .

Let be a finite nonatomic and complete measure space. Denote by and the right derivative of and , respectively. We define It is well known that the Orlicz function space is a Banach space when it is equipped with the Luxemburg norm or equipped with the Amemiya-Orlicz norm , denote Orlicz function spaces equipped with the Luxemburg norm. , denote Orlicz function spaces equipped with the Orlicz norm. It is well known that , .

Definition 14 (see [13]). We say that an -function if there exist and such that whenever .

Lemma 15 (see [13, 14]). Let be Orlicz function space. Then the following statements are equivalent:(1)Every hyperplane of is approximatively compact;(2) is smooth space and .

Lemma 16 (see [15]). () has the Radon-Nikodym property if and only if .

Lemma 17 (see [15]). is a strictly convex space if and only if is strictly convex.

Lemma 18 (see [15]). is reflexive if and only if .

Theorem 19. If every hyperplane of is approximatively compact, then .

Proof. Since every hyperplane of is approximatively compact, we obtain that has the Radon-Nikodym property (see [14]). Therefore, by Lemma 16, we have . This completes the proof.

Example 20. If is strictly convex and , then is not nearly dentable. In fact, suppose that is nearly dentable. Since is strictly convex, we obtain that is a strictly convex space by Lemma 17. Since , we obtain that is a smooth space. This implies that if is a separable closed subspace of , then is a singleton. Therefore, by Theorem 10, we obtain that has the Radon-Nikodym property. Then, by Lemma 16, we obtain that , a contradiction.

Example 21. If is strictly convex, and , then is nearly dentable not nearly dentable. In fact, by Theorem 19, we obtain that every hyperplane of is approximatively compact. Therefore, by Theorem 11, we obtain that is nearly dentable. Since , we obtain that is not reflexive. This implies that is not nearly dentable.

5. Applications to Theory of Generalized Inverses of Banach Space

Let be a linear bounded operator from into . Let and denote the domain, range, and null space of , respectively. If or , the operator equation is generally ill-posed. In applications, one usually looks for the best approximative solution (b.a.s.) to the equation (see [16]).

A point is called the best approximative solution to the operator equation , if where (see [16]). Nashed and Votruba [16] introduced the concept of the (set-valued) metric generalized inverse .

Definition 22. Let be Banach spaces and let be a linear operator from to . The set-valued mapping defined by for any is called the (set-valued) metric generalized inverse of , where

Definition 23. is said to be Hausdorff distance of and if

During the last three decades, the linear generalized inverses of linear operators in Banach spaces and their applications have been investigated by many authors. In 2008, Chen et al. [8], under some assumption on the geometric structure of Banach space, gave the criteria for the existence of continuous metric generalized inverse. In 2005, Wei et al. [17] presented componentwise condition numbers for the problems of Moore-Penrose generalized matrix inverses and linear least squares. For the selection of the metric generalized inverse of linear operator in Banach space, in 2008, Hudzik et al. [18] obtained a bounded homogeneous selection.

Theorem 24. Let be a nearly dentable space, a nearly dentable space, a one-dimensional subspace of , and a closed semi-Chebychev set. Then . Moreover, if for any there exists such that is compact and , then(1) is compact and is upper semicontinuous at ;(2) is lower semicontinuous at if and only if the function is lower semicontinuous at .

Proof. Since is closed, we obtain that by the closed range theorem. Since dim, there exists such that . This implies that is a hyperplane of . By proof of Theorem 2.1 of [13], we obtain that is a proximinal set. Since is a semi-Chebychev set, we obtain that is a Chebychev set. Let . Then there exists a unique such that . Hence there exists such that . Since is a nearly dentable space, we obtain that is reflexive. This implies that is a proximinal set. Hence we obtain that .
We will prove that the set is compact. Let . Since is a nearly dentable space and is a hyperplane of , by Theorem 12, we obtain that metric projector operator is upper semicontinuous. Moreover, since is a Chebychev set, we obtain that metric projector operator is continuous; that is, as . Since is linear bounded operator, we obtain that is a closed subspace. Put Then, by the inverse operator theorem, we obtain that This implies that as . Since, for any , there exists such that , we have . Moreover, we may assume without loss of generality that as . Therefore, by the definition of set-valued metric generalized inverse, we obtain that Suppose that is not compact. Then there exists a sequence such that has no Cauchy subsequence. Since is a nearly dentable space, we obtain that is reflexive. Hence we may assume without loss of generality that . Noticing that is a closed convex set, we obtain that is a weakly closed convex set. Then . Put Noticing that , , and , we obtain that . By , we obtain that . Moreover, it is easy to see that . Then there exists such that is compact and . We will derive a contradiction for each of the following two cases.
Case I. Let be a finite set. Then we may assume without loss of generality that . Hence, for any , there exists such that , where . Hence there exists an open set such that . Since is a nearly dentable space, we obtain that . By the separation theorem, there exists such that Noticing that and , we obtain that , which contradict as . This implies that the sequence has a Cauchy subsequence.
Case II. Let be an infinite set. Then there exists a subsequence of such that . Since is a compact set, we obtain that is relatively compact. This implies that has a Cauchy subsequence. Hence is compact. By , we obtain that is compact.
We next will prove that the set-valued mapping is upper semicontinuous at . Let be an open set and . We claim that there exists such that Otherwise, there exists a sequence such that . Let be a Cauchy sequence. Noticing that , we have as . Then Hence there exists such that . Since , there exists such that Hence, if , then This means that if , then , a contradiction. Hence there exists such that Let Since is a nearly dentable space, we obtain that the metric projector operator is upper semicontinuous. Since metric projector operator is upper semicontinuous and , there exists such that whenever . This means that whenever . Moreover, by , there exists such that whenever . Let . We claim that whenever . In fact, if , then there exists such that . Let . Then and . Since , there exists such that . Then This implies that . Then . This implies that . Therefore, by formula (59), we have .
Let be an open set and . Then, by formula (59), we have . Hence there exists natural number such that whenever . This implies that whenever . Hence we obtain that the set-valued mapping is upper semicontinuous at .
Let be lower semicontinuous at . It is easy to see that . Suppose that is not lower semicontinuous at . Then there exist a sequence and such that and Since is lower semicontinuous at , we obtain that, for any , there exists such that as . This implies that for any , we have We claim that In fact, suppose that formula (71) is not true. Since is a compact set, there exist , , and such that Since is compact, we may assume that as . Therefore, by formula (72), we have . Moreover, by formula (71), we obtain that a contradiction. Hence formula (71) is true.
Since is upper semicontinuous at , we obtain that for any , there exists such that whenever . This implies that for any , we obtain that dist as . Hence, for any , we obtain that We claim that Otherwise, there exist , , and such that Moreover, we have a contradiction. This implies that formula (76) is true. Therefore, by formulas (71) and (76), we have which contradict formula (69). Hence This implies that the function is lower semicontinuous at .
On the other hand, if is lower semicontinuous at , then . Hence, if , then Hence, for any , there exists such that This implies that is lower semicontinuous at . This completes the proof.

Theorem 25. Let be a nearly dentable space, approximatively compact, a one-dimensional subspace, and a closed semi-Chebychev set. Then(1) and is upper semicontinuous;(2) is lower semicontinuous at if and only if the function is lower semicontinuous at .

Proof. Let . Since is approximatively compact, we obtain that, for any , there exists such that is compact and . Moreover, since is approximatively compact, we obtain that is a nearly dentable space. By Theorem 24, Theorem 25 is true, which completes the proof.

Conflict of Interests

The authors declare that there is no conflict of interests regarding the publication of this paper.

Acknowledgments

This work is supported by “China Natural Science Fund under Grant 11401084 and Scientific Research Foundation for the Rturned Overseas Chinese Scholars of Heilongjiang Province (Grant no. LC201402)”.